A First Course on Kinetics and Reaction Engineering Class 35 on - - PowerPoint PPT Presentation

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A First Course on Kinetics and Reaction Engineering Class 35 on - - PowerPoint PPT Presentation

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A First Course on Kinetics and Reaction Engineering Class 35 on Unit 33 Where Were Going Part I - Chemical Reactions Part II - Chemical Reaction Kinetics Part III - Chemical Reaction Engineering A. Ideal Reactors B.

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SLIDE 1

A First Course on Kinetics and Reaction Engineering

Class 35 on Unit 33

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SLIDE 2

Where We’re Going

  • Part I - Chemical Reactions
  • Part II - Chemical Reaction Kinetics
  • Part III - Chemical Reaction Engineering
  • A. Ideal Reactors
  • B. Perfectly Mixed Batch Reactors
  • C. Continuous Flow Stirred Tank Reactors
  • D. Plug Flow Reactors
  • E. Matching Reactors to Reactions
  • Part IV - Non-Ideal Reactions and Reactors
  • A. Alternatives to the Ideal Reactor Models
  • 33. Axial Dispersion Model
  • 34. 2-D and 3-D Tubular Reactor Models
  • 35. Zoned Reactor Models
  • 36. Segregated Flow Model
  • 37. Overview of Multi-Phase Reactors
  • B. Coupled Chemical and Physical Kinetics

2

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SLIDE 3

Axial Dispersion Model

  • The axial dispersion model is a modified form of the plug flow reactor

model

  • it adds a diffusion-like term that accounts for mixing in the axial direction
  • the quantity Dax is not a diffusion coefficient
  • it is called the dispersion coefficient
  • it has one value that is used for all species being modeled
  • it can be found
  • from correlations of the axial Peclet number (or its inverse, the dispersion number) as

a function of the Reynolds number

  • experimentally, for example using stimulus and response measurements involving a

tracer (as is done when measuring the age function

  • In commercial scale packed bed reactors, including axial dispersion

doesn’t affect conversion as long as the reactor length is greater than ca. 50 catalyst particle diameters

  • The axial dispersion model can still be used to empirically model a reactor
  • When Dax = 0, it is a PFR; as Dax becomes very large, it approaches a CSTR

−Dax d 2Ci dz2 + d dz usCi

( ) =

νi, jrj

j=all reactions

∑

Peax = ul Dax

3

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SLIDE 4

Boundary Conditions and Numerical Solution

  • The axial dispersion model is a second order differential equation
  • There are several ways the boundary conditions could be formulated
  • Danckwerts boundary conditions are very common
  • First Danckwerts boundary condition recognizes that as soon as the feed

enters the reactor, it will be diluted by axial mixing

  • Much like what happens to feed entering a CSTR
  • It requires that the flow of A upstream of the reactor entrance (where no mixing has occurred)

must be equal to the net flow of A (due to convection and mixing) at the inlet to the reactor

  • Second Danckwerts boundary condition simply requires that the

concentration stops changing at the point where the flow leaves the reactor.

  • The axial dispersion model involves mixed boundary value ordinary

differential equations

  • Numerical solution requires different methods than are used for initial value problems
  • Discussed in Supplemental Unit S6

at z = 0; uCA 0

( )− Dax

dCA dz

z=0

= u0CA, feed at z = L; dCA dz

z=L

= 0

4

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SLIDE 5

Questions?

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SLIDE 6

Suppose liquid phase reaction (1) is second order in the concentration of A. The reaction proceeds isothermally with a rate coefficient equal to 0.05 dm3 mol-1 min-1. The liquid density is constant. The feed to the process is taken from a tank containing A at a concentration of 1 M and enters the reactor with a superficial velocity of 6.0 dm min-1. Plot the concentration of A versus axial position within a tubular reactor for axial dispersion coefficients of 0.1, 1, 10, 100 and 1000 dm2 min-1. A → B (1) Read through the problem statement. Each time you encounter a quantity, write it down and equate it to the appropriate variable. When you have completed doing so, if there are any additional constant quantities that you know will be needed and that can be calculated from the values you found, write the equations needed for doing so.

Activity 33.1

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SLIDE 7

Suppose liquid phase reaction (1) is second order in the concentration of A. The reaction proceeds isothermally with a rate coefficient equal to 0.05 dm3 mol-1 min-1. The liquid density is constant. The feed to the process is taken from a tank containing A at a concentration of 1 M and enters the reactor with a superficial velocity of 6.0 dm min-1. Plot the concentration of A versus axial position within a tubular reactor for axial dispersion coefficients of 0.1, 1, 10, 100 and 1000 dm2 min-1. A → B (1) Read through the problem statement. Each time you encounter a quantity, write it down and equate it to the appropriate variable. When you have completed doing so, if there are any additional constant quantities that you know will be needed and that can be calculated from the values you found, write the equations needed for doing so. k = 0.05 dm3 mol-1 min-1 CA,feed = 1 mol L-1 us = 6.0 dm min-1 Dax = 0.1, 1, 10, 100 or 1000 dm2 min-1

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SLIDE 8

Suppose liquid phase reaction (1) is second order in the concentration of A. The reaction proceeds isothermally with a rate coefficient equal to 0.05 dm3 mol-1 min-1. The liquid density is constant. The feed to the process is taken from a tank containing A at a concentration of 1 M and enters the reactor with a superficial velocity of 6.0 dm min-1. Plot the concentration of A versus axial position within a tubular reactor for axial dispersion coefficients of 0.1, 1, 10, 100 and 1000 dm2 min-1. A → B (1) Generate the mole balance design equations needed to model the axially- dispersed tubular reactor by simplification of the general design equations found in Unit 33 or on the AFCoKaRE Exam Handout.

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SLIDE 9

Suppose liquid phase reaction (1) is second order in the concentration of A. The reaction proceeds isothermally with a rate coefficient equal to 0.05 dm3 mol-1 min-1. The liquid density is constant. The feed to the process is taken from a tank containing A at a concentration of 1 M and enters the reactor with a superficial velocity of 6.0 dm min-1. Plot the concentration of A versus axial position within a tubular reactor for axial dispersion coefficients of 0.1, 1, 10, 100 and 1000 dm2 min-1. A → B (1) Generate the mole balance design equations needed to model the axially- dispersed tubular reactor by simplification of the general design equations found in Unit 33 or on the AFCoKaRE Exam Handout.

−Dax d 2Ci dz2 + d dz usCi

( ) =

νi, jrj

j=all reactions

∑

at z = 0; usCi z = 0

( )− Dax

dCi dz

z=0

= usCi, feed at z = L; dCi dz

z=L

= 0 −Dax d 2CA dz2 + us dCA dz = −r usCA 0

( )− Dax

dCA dz

z=0

− usCA, feed = 0 dCi dz

z=L

= 0

9

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SLIDE 10
  • Identify the independent and dependent variables, if appropriate, and the

unknown quantities to be found by solving the equations

−Dax d 2CA dz2 + us dCA dz = −r usCA 0

( )− Dax

dCA dz

z=0

− usCA, feed = 0 dCi dz

z=L

= 0

10

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SLIDE 11
  • Identify the independent and dependent variables, if appropriate, and the

unknown quantities to be found by solving the equations

  • Independent variable: z
  • Dependent variable: CA
  • Solution will give CA(z) between z = 0 and z = L
  • Assuming that the equations will be solved numerically, specify the

information that must be provided and show how to calculate any unknown values

11

−Dax d 2CA dz2 + us dCA dz = −r usCA 0

( )− Dax

dCA dz

z=0

− usCA, feed = 0 dCi dz

z=L

= 0

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SLIDE 12
  • Identify the independent and dependent variables, if appropriate, and the

unknown quantities to be found by solving the equations

  • Independent variable: z
  • Dependent variable: CA
  • Solution will give CA(z) between z = 0 and z = L
  • Assuming that the equations will be solved numerically, specify the

information that must be provided and show how to calculate any unknown values

  • Must supply code that is given CA and its derivatives as a function of z and uses them to

evaluate the ODEs

  • Dax and us are known constants
  • r = k(CA)2
  • k is a known constant
  • Must supply code that evaluates the boundary conditions given values of CA, z and the

derivatives at z = 0 and z = L

  • Dax, CA,feed, us and L are known constants
  • Identify what variables will become known upon solving the design

equations and show how those variables can be used to answer the questions that were asked in the problem

12

−Dax d 2CA dz2 + us dCA dz = −r usCA 0

( )− Dax

dCA dz

z=0

− usCA, feed = 0 dCi dz

z=L

= 0

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SLIDE 13
  • Identify the independent and dependent variables, if appropriate, and the

unknown quantities to be found by solving the equations

  • Independent variable: z
  • Dependent variable: CA
  • Solution will give CA(z) between z = 0 and z = L
  • Assuming that the equations will be solved numerically, specify the

information that must be provided and show how to calculate any unknown values

  • Must supply code that is given CA and z and uses them to evaluate all other quantities in the

ODE except derivatives

  • Dax and us are known constants
  • r = k(CA)2
  • k is a known constant
  • Must supply code that evaluates the boundary conditions given values of CA, z and the

derivatives at z = 0 and z = L

  • Dax, CA,feed, and us are known constants
  • Identify what variables will become known upon solving the design

equations and show how those variables can be used to answer the questions that were asked in the problem

  • CA(z) between z = 0 and z = L will be known, can use it to make the requested plot

13

−Dax d 2CA dz2 + us dCA dz = −r usCA 0

( )− Dax

dCA dz

z=0

− usCA, feed = 0 dCi dz

z=L

= 0

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SLIDE 14
  • The results are shown in this

plot

  • If Dax = 0, the reactor is a

PFR

  • What would the plot look like?
  • If Dax = ∞, the reactor is a

CSTR

  • What would the plot look like?
  • Do the results at

intermediate Dax make sense physically?

Axial Position (m)

50 100 150 200 250 300 350 400

Concentration of A (M)

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Dax = 0.1 Dax = 1 Dax = 10 Dax = 100 Dax = 1000

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SLIDE 15

Numerical Solution of Boundary Value ODEs

  • Identification of boundary value ordinary differential equations (ODEs)
  • No partial derivatives
  • All ordinary derivatives with respect to the same independent variable
  • Independent variable ranges from some known lower limit to a different upper limit
  • some boundary conditions apply at the lower limit and other boundary conditions apply at

the upper limit

  • Requirements imposed by the MATLAB solver, bvp4c
  • Higher order ODEs reduced to sets of first order ODEs
  • Each ODE written in the form
  • Sij equal zero or constant (here, Sij is equal to zero)
  • To convert any second order ODE into a set of two first order ODEs
  • Define a new variable that is equal to the first derivative of the original dependent variable with

respect to the independent variable

  • This definition becomes one of the two first order ODEs in the set
  • In the original second order ODE
  • Replace any occurrence of the second derivative of the original dependent variable with

the first derivative of the new variable

  • Replace any occurrence of the first derivative of the original dependent variable with the

new variable

  • In the boundary conditions, replace the first derivative of the original dependent variable with

the new variable

dyi dx = Sij yj x

all j

∑

+ fi x, y

( )

15

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SLIDE 16

Continuing the Present Problem

  • To solve numerically we need to convert the second order ODE to two first
  • rder ODEs
  • Let
  • This becomes one of the two first order ODEs
  • Note that
  • Substitute and for and in the original ODE
  • This becomes the second of the two first order ODEs
  • Also re-write the boundary conditions replacing with

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−Dax d 2CA dz2 + us dCA dz = −r Y = dCA dz dY dz = d 2CA dz2 dY dz Y d 2CA dz2 dCA dz −Dax dY dz + usY = −r dCA dz Y usCA 0

( )− Dax

dCA dz

z=0

− usCA, feed = 0 ⇒ usCA 0

( )− DaxY 0 ( )− usCA, feed = 0

dCi dz

z=L

= 0 ⇒ Y L

( ) = 0

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SLIDE 17

Choosing the MATLAB Template File

  • The equations are boundary value ODEs; they do not contain a singularity

at x = 0

  • The MATLAB template file SolvBVDif.m can be used to solve them
  • Follow the step by step instructions provided with this Supplemental Unit S6
  • Save a copy of SolvBVDif.m as S6_Example_1.m
  • Change the initial comment to reflect the purpose of the modified file
  • Change the function declaration to match the filename, but without the “.m”
  • % Modified version of the MATLAB template file SolvBVDif.m that was used to

% solve the in-class problem in Class 35 of CE 329, Fall 2014. % function result = S6_Example_1 % Known quantities and constants (in consistent units) Dax = 0.1; us = 0.6.0; k = 0.05; CA0 = 1.0; L = 400; % May need to change this; just a guess 17

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SLIDE 18

Second Required Modification

% Function that evaluates the derivatives function dydx = bvodes(x,y) dydx = [ y(2) (1/Dax)*(k*y(1)^2 - us*y(2)) ]; end % of internal function bvodes % Function that evaluates the derivatives function dydx = bvodes(x,y) % EDIT HERE (Required modification 2 of 6): dydx = [ % Evaluate dy1/dx here % Evaluate dy2/dx here % and so on, one dyi/dx per line ]; end % of internal function bvodes 18

dCA dz = Y dY dz = 1 Dax r − usY

( )

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SLIDE 19

% Function that calculates the errors at the boundaries function res = bvs(y_at_start,y_at_end) % y_at_start is a column vector containing y values at the starting % x boundary and y_at_end is a column vector containing y values at % the ending x boundary. res = [ us*y_at_start(1)-Dax*y_at_start(2)-us*CA0 y_at_end(2) ]; end % of internal function bvs

Third Required Modification

% Function that calculates the errors at the boundaries function res = bvs(y_at_start,y_at_end) % y_at_start is a column vector containing y values at the starting % x boundary and y_at_end is a column vector containing y values at % the ending x boundary. % EDIT HERE (Required modification 3 of 6): res = [ % Calculate the error in the first boundary condition here % Calculate the error in second boundary condition here % and so on, one boundary condition per line ];

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usCA 0

( )− DaxY 0 ( )− usCA, feed = 0

Y L

( ) = 0

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SLIDE 20

Fourth Required Modification

% EDIT HERE (Required modification 4 of 6): % Set up the initial mesh x_range_low = % insert starting value of the valid range of x here x_range_high = % insert ending value of the valid range of x here n_mesh_points = % insert the number of mesh points here % Set up the initial mesh x_range_low = 0.0; x_range_high = L; n_mesh_points = 40;

0 ≤ x ≤ L

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SLIDE 21

% Guesses yinit = [ CA0

  • CA0/L

];

Fifth Required Modification

% Guesses % EDIT HERE (Required modification 5 of 6): yinit = [ % Enter a guess for y1 (one value for the whole x range) here % Enter a guess for y2 (one value for the whole x range) here % and so on, one guess per line ]; 21

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SLIDE 22

Sixth and Final Required Modification

% EDIT HERE (Required modification 6 of 6): % Calculate any other desired quantities from the results, noting % result.x(i) contains the final x value of mesh point i % result.y(i,j) contains the final value of yj at mesh point i

  • Here, just use results to plot CA vs z

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SLIDE 23

Where We’re Going

  • Part I - Chemical Reactions
  • Part II - Chemical Reaction Kinetics
  • Part III - Chemical Reaction Engineering
  • A. Ideal Reactors
  • B. Perfectly Mixed Batch Reactors
  • C. Continuous Flow Stirred Tank Reactors
  • D. Plug Flow Reactors
  • E. Matching Reactors to Reactions
  • Part IV - Non-Ideal Reactions and Reactors
  • A. Alternatives to the Ideal Reactor Models
  • 33. Axial Dispersion Model
  • 34. 2-D and 3-D Tubular Reactor Models
  • 35. Zoned Reactor Models
  • 36. Segregated Flow Model
  • 37. Overview of Multi-Phase Reactors
  • B. Coupled Chemical and Physical Kinetics

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